Miami Dolphins running back Jay Ajayi ran for 204 yards with two touchdowns on Sunday against the Pittsburgh Steelers, the first 200-yard rushing performance in the NFL this year. The Dolphins won the game, in no small part because of Ajayi, and now, the NFL has recognized the second-year running back’s performance.
The NFL has named Ajayi’s the AFC Offensive Player of the Week for Week 6.
Ajayi averaged 8.2 yards per attempt on 25 carries during the game. He solidified Miami’s win with a 62-yard touchdown to extend Miami’s lead to 30-15 just 12 seconds after the Steelers had scored a touchdown of their own.
Ajayi is the first Dolphins player to receive a Player of the Week award this season. The last running back to win the Offensive Player of the Week award for the Dolphins was Reggie Bush in 2012. He is the ninth running back in team history to win the award, joining Bush (twice), Ronnie Brown, Ricky Williams (three times), Lamar Smith (twice), Karim Abdul-Jabbar, Bernie Parmalee, Smamie Smith, and Troy Stadford.
The Dolphins won three AFC Player of the Week awards last year, with Jarvis Landry earning the special teams award in Week 1, Cameron Wake the defensive award in Week 6, and Ryan Tannehill the offensive award in Week 7.
Ajayi is also nominated for the FedEx Ground Player of the Week, awarded based on a fan vote. Fans can submit their vote for Ajayi by going to the NFL’s Twitter page (@NFL) and tweeting Jay Ajayi with the hashtag #AirAndGround.